Question

Question #445f7

Answer 1

`5x^2-10x-2=0`

Explanation:

If the equation is `ax^2+bx+c=0` then alpha+beta= b/a and alpha.beta=c/a.
Comparing `ax^2+bx+c=0, x^2+(-2)x+5=0` we get alpha+beta = -2 and alpha.beta=5.
Now, 1/alpha + 1/beta =(alpha+beta)/alpha.beta= -2/5.
Hence new equation is:
`x^2+(-2)x+(-2/5)=0`
Or,`5x^2-10x-2=0`

Answer 2

see a solution process below;

Explanation:

Let `A` and `B` be alpha and beta respectively..

`x^2-2x + 5 = x^2 -(A+B)x + AB`

`A + B = 2`

`AB = 5`

When `A + B` and `1/A+1/B` are the roots of the equation,

The expression becomes;

`x^2-(A+B+(1/A+1/B)x+[(A+B)(1/A+1/B)]`

Therefore;

`x^2-(A+B+((A+B)/(AB))x+[(A+B)((A+B)/(AB))]`

Substituting the values;

`x^2-(2+2/5)x+(2*2/5)`

`x^2-(12/5)x+4/5`

Multiply through by `5`

`5x^2 - 12x + 4`

Hope this helps!